Answer
Verified
375.6k+ views
Hint :The $ N{a_2}C{O_3} $ is sodium carbonate, which is being converted in sodium bicarbonate $ \left( {NaHC{O_3}} \right). $ $ A{l_2}{\left( {S{O_4}} \right)_3} $ is aluminum sulphate, and is being converted to aluminum hydroxide $ \left( {Al{{\left( {OH} \right)}_3}} \right). $ $ N{a_2}{S_2}{O_3} $ is sodium thiosulphate is converted to tetrathionate anion $ \left( {{S_4}O_6^{2 - }} \right). $
Complete Step By Step Answer:
We have:
(i) $ N{a_2}C{O_3} \to NaHC{O_3} $
For $ N{a_2}C{O_3} \to NaHC{O_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ N{a_2}C{O_3}, $ carbonate anion $ \left( {CO_3^{2 - }} \right) $ has $ - 2 $ charge. This means that the total positive charge contributed by the cations $ \left( {2N{a^ + }} \right) $ is $ + 2. $ In $ NaHC{O_3}, $ bicarbonate anion has $ - 1 $ charge, and hence total positive charge contributed by cation $ \left( {N{a^ + }} \right) $ is $ + 1. $
Therefore, n-factor value $ = 2 - 1 = 1 $
(ii) $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $
For $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ A{l_2}{\left( {S{O_4}} \right)_3}, $ sulphate anion $ \left( {SO_4^{2 - }} \right) $ has $ - 2 $ charge. The total negative charge will therefore be $ - 2 \times 3 = 6 $ as there are three sulphate ions. This means that the total positive charge contributed by the cations $ \left( {2A{l^{3 + }}} \right) $ is $ + 6. $ In $ Al{\left( {OH} \right)_3}, $ hydroxide ion has $ - 1 $ charge, and hence three hydroxide ions contribute $ - 3 $ charge. The total positive charge contributed by cation $ \left( {A{l^{3 + }}} \right) $ is $ + 3. $
Therefore, n-factor value $ = 6 - 3 = 3 $
(iii) $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $
For $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $ the n-factor is the product of change in oxidation state and number of atoms undergoing change. Since the change undergone is as follows: $ {S_2}O_3^{2 - } \to {S_4}O_6^{2 - } $ we need to calculate the oxidation state of sulphur atom in each of these. The oxidation states of sulphur in $ {S_2}O_3^{2 - } $ and $ {S_4}O_6^{2 - } $ are $ 2 $ and $ 2.5 $ respectively and hence the change in oxidation state is $ 2.5 - 2 = 0.5. $ The number of sulphur atoms undergoing change are $ 4 - 2 = 2. $
Therefore, n-factor value $ = 0.5 \times 2 = 1 $
Hence, the required answer is (B) $ 1,3,1. $
Note :
The n-factor can be calculated for different compounds according to their characteristics: whether they are an acid, base, salt or if the reaction undergone by the compound is a redox reaction. For acids, the n-factor is represented in the form of basicity, while for bases, the n-factor is represented in the form of acidity. For salts, the number of positive or negative ions are counted. For redox reactions, the change in oxidation number or change in their reduction number in both sides of the reaction is considered.
Complete Step By Step Answer:
We have:
(i) $ N{a_2}C{O_3} \to NaHC{O_3} $
For $ N{a_2}C{O_3} \to NaHC{O_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ N{a_2}C{O_3}, $ carbonate anion $ \left( {CO_3^{2 - }} \right) $ has $ - 2 $ charge. This means that the total positive charge contributed by the cations $ \left( {2N{a^ + }} \right) $ is $ + 2. $ In $ NaHC{O_3}, $ bicarbonate anion has $ - 1 $ charge, and hence total positive charge contributed by cation $ \left( {N{a^ + }} \right) $ is $ + 1. $
Therefore, n-factor value $ = 2 - 1 = 1 $
(ii) $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $
For $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ A{l_2}{\left( {S{O_4}} \right)_3}, $ sulphate anion $ \left( {SO_4^{2 - }} \right) $ has $ - 2 $ charge. The total negative charge will therefore be $ - 2 \times 3 = 6 $ as there are three sulphate ions. This means that the total positive charge contributed by the cations $ \left( {2A{l^{3 + }}} \right) $ is $ + 6. $ In $ Al{\left( {OH} \right)_3}, $ hydroxide ion has $ - 1 $ charge, and hence three hydroxide ions contribute $ - 3 $ charge. The total positive charge contributed by cation $ \left( {A{l^{3 + }}} \right) $ is $ + 3. $
Therefore, n-factor value $ = 6 - 3 = 3 $
(iii) $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $
For $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $ the n-factor is the product of change in oxidation state and number of atoms undergoing change. Since the change undergone is as follows: $ {S_2}O_3^{2 - } \to {S_4}O_6^{2 - } $ we need to calculate the oxidation state of sulphur atom in each of these. The oxidation states of sulphur in $ {S_2}O_3^{2 - } $ and $ {S_4}O_6^{2 - } $ are $ 2 $ and $ 2.5 $ respectively and hence the change in oxidation state is $ 2.5 - 2 = 0.5. $ The number of sulphur atoms undergoing change are $ 4 - 2 = 2. $
Therefore, n-factor value $ = 0.5 \times 2 = 1 $
Hence, the required answer is (B) $ 1,3,1. $
Note :
The n-factor can be calculated for different compounds according to their characteristics: whether they are an acid, base, salt or if the reaction undergone by the compound is a redox reaction. For acids, the n-factor is represented in the form of basicity, while for bases, the n-factor is represented in the form of acidity. For salts, the number of positive or negative ions are counted. For redox reactions, the change in oxidation number or change in their reduction number in both sides of the reaction is considered.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE